I'm fairly new to Python and am trying to find a simple way to round
floats to a specific number of significant digits. I found an old
post on this list with exactly the same problem:
<http://mail.python.org/pipermail/tutor/2004-July/030268.html>
Is there something (a function?) in Python 2.3.4 that will round a
result to n significant digits, or do I need to roll my own? I don't
see one in the math module.
I mean something like rounding(float, n) that would do this:
float = 123.456789, n = 4, returns 123.5
float = .000000123456789, n = 2, returns .00000012
float = 123456789, n = 5, returns 123460000
Thanks,
Dick Moores
And another post gave this solution:
<http://mail.python.org/pipermail/tutor/2004-July/030311.html>
I expect the easiest way to do this in Python is to convert to
string using an %e format, then convert that back to float again.
Like this:
def round_to_n(x, n):
if n < 1:
raise ValueError("number of significant digits must be >= 1")
# Use %e format to get the n most significant digits, as a string.
format = "%." + str(n-1) + "e"
as_string = format % x
return float(as_string)
Converting to a string seemed like an awkward hack to me, so I came up
with a mathematical solution:
import math
def round_figures(x, n):
"""Returns x rounded to n significant figures."""
return round(x, int(n - math.ceil(math.log10(abs(x)))))
print round_figures(123.456789,4)
print round_figures(.000000123456789,2)
print round_figures(123456789,5)
print round_figures(0.987,3)
print round_figures(0.987,2)
print round_figures(0.987,1)
print round_figures(-0.002468,2)
123.5
1.2e-07
123460000.0
0.987
0.99
1.0
-0.0025
Since the built-in round(x,n) can do rounding in the 10's and 100's
places just as easy as 0.1's and 0.01's, my function just counts how
many digits are in use and rounds off n digits away.
I thought others might find this solution useful. Or somebody else
might share a nicer way.
Richard Wagner
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